differential equations Solution b> According to Eulers method the approximate value of f at x=a with a step size h is f(a) = f(a-h) + f\'(a-h)*h we are given f(0) = 1 => f\'(x) = 5y-4x f\'(0) = 5(1) - 4(0) =5 f(.1) = f(0) + f\'(0)*(.1) = 1 + 5*.1 = 1.5 => f(0.2) =f(.1) +f\'(.1)*(.1) now when x = .1 , f = 1.5 => f\'(.1) = 5(1.5) - 4(.1) = 7.1 => f(0.2) =1.5 + (7.1)*(.1) = 2.21.

1. differential equations
Solution
b>
According to Eulers method
the approximate value of f at x=a with a step size h is
f(a) = f(a-h) + f'(a-h)*h
we are given f(0) = 1
=> f'(x) = 5y-4x
f'(0) = 5(1) - 4(0) =5
f(.1) = f(0) + f'(0)*(.1)
= 1 + 5*.1 = 1.5
=> f(0.2) =f(.1) +f'(.1)*(.1)
now when x = .1 , f = 1.5
=> f'(.1) = 5(1.5) - 4(.1) = 7.1
=> f(0.2) =1.5 + (7.1)*(.1) = 2.21